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return a procedure from make-variable-like-transformer

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So that make-variable-like-transformer produces a value that passes both procedure? and set!-transformer?
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AlexKnauth 2019-03-17 22:54:11 -04:00 committed by Ryan Culpepper
parent 1efe1ebbf1
commit 780647f0eb
1 changed files with 9 additions and 4 deletions
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13
racket/collects/syntax/transformer.rkt
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@ -4,17 +4,22 @@
(provide make-variable-like-transformer) (provide make-variable-like-transformer)
(struct variable-like-transformer [procedure]
#:property prop:procedure (struct-field-index procedure)
#:property prop:set!-transformer (struct-field-index procedure))
(define (make-variable-like-transformer ref-stx [set!-handler #f]) (define (make-variable-like-transformer ref-stx [set!-handler #f])
(unless (syntax? ref-stx) (unless (or (syntax? ref-stx) (procedure? ref-stx))
(raise-type-error 'make-variable-like-transformer "syntax?" ref-stx)) (raise-type-error 'make-variable-like-transformer "(or/c syntax? procedure?)" ref-stx))
(unless (or (syntax? set!-handler) (procedure? set!-handler) (eq? set!-handler #f)) (unless (or (syntax? set!-handler) (procedure? set!-handler) (eq? set!-handler #f))
(raise-type-error 'make-variable-like-transformer "(or/c syntax? procedure? #f)" set!-handler)) (raise-type-error 'make-variable-like-transformer "(or/c syntax? procedure? #f)" set!-handler))
(make-set!-transformer (variable-like-transformer
(lambda (stx) (lambda (stx)
(syntax-case stx (set!) (syntax-case stx (set!)
[id [id
(identifier? #'id) (identifier? #'id)
ref-stx] (cond [(procedure? ref-stx) (ref-stx stx)]
[else ref-stx])]
[(set! id val) [(set! id val)
(cond [(procedure? set!-handler) (cond [(procedure? set!-handler)
(set!-handler stx)] (set!-handler stx)]